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Quantum Perturbation Theory

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the first excited state of the Hydrogen atom. The principle quantum number is given by n = 2 and so it is four-fold degenerate. Consider now a weak perturbation in the form of V = λxy, where x and y are the Cartesian coordinates of the electron with respect to the proton. The explicit forms of the involved spherical harmonics are also given below:

    [tex]Y_{00} = \frac{1}{\sqrt{4\pi}}[/tex]

    [tex]Y_{10} = \frac{3}{\sqrt{4\pi}}cos(\theta) = \frac{3}{\sqrt{4\pi}}\frac{z}{r}[/tex]

    [tex]Y_{11} = -\frac{3}{\sqrt{8\pi}}sin(\theta)e^{i\phi} = -\frac{3}{\sqrt{8\pi}}\frac{x + iy}{r}[/tex]

    [tex]Y_{1-1} = \frac{3}{\sqrt{8\pi}}sin(\theta)e^{-i\phi} = \frac{3}{\sqrt{8\pi}}\frac{x - iy}{r}[/tex]​

    Based on the first-order degenerate perturbation theory, describe qualitatively how the degenerate levels will shift or split due to the perturbation.

    2. The attempt at a solution

    In this case of four-fold degeneracy, the first-order corrections of energy E1 are given by the eigenvalues of a W-matrix, which has the following elements:

    Wij = < ψi0|V| ψjo>, i,j = 1,2,3,4​

    Here for simplicity I denote ψ200, ψ21-1, ψ210, ψ211 using ψ1, ψ2, ψ3, ψ4 respectively, and ψnlm stands for the state of quantum number n, l, and m.

    Then I use the given spherical harmonics to determine whether certain element of matrix W is zero. For example,

    W11 = <ψ10|V|ψ10> = 0​

    because the φ part of above equation converges to 0 when integrated from 0 to 2π.

    After I went through all the elements of W, I got:

    W11 = W22 = W33 = W44 = W13 = W23 = W43 = W31 = W32 = W34 = 0​

    Since I am asked to give a qualitative description, I did not work out the specific values of the other elements, but denote them using:

    W12 = a1, W14 = b1, W21 = a2, W24 = c1, W41 = b2, W42 = c2

    So finally I have this characteristic equation for W:

    λ4- (a1a2 - b1b2 + c1c22 - (a1b2c1 + a2b1c2)λ = 0​

    Here λ represents the eigenvalues of W and hence is the first-order correction of energy.

    Thanks to phyzguy. I lost the root λ = 0 during my previous calculation.

    There are totally four degenerate states in the beginning, and only three different energy levels after perturbation. It is said that "typically the perturbation will break or lift the degeneracy", and I do not understand why here I lost one degeneracy. So I was wondering whether my calculations were wrong somewhere, i.e., some of the elements of W may not be zero.

    I have been working on this problem for days, calculating around, but still cannot figure it out.

    Last edited: Aug 29, 2010
  2. jcsd
  3. Aug 29, 2010 #2


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    The characteristic equation of an NxN matrix is always N-dimensional and has N roots. The problem is that you have dropped a factor of lambda in your characteristic equation. it should read:
    [tex]\lambda^4 - (a1a2 - b1b2 + c1c2)\lambda^2 - (a1b2c1 + a2b1c2)\lambda = 0[/tex]
    This means that there is an additional root of lambda=0. This means that one of the energy levels (number 3 in this case) is not impacted by the perturbation.
  4. Aug 29, 2010 #3
    Thanks very much for ur reply and the solution is pretty clear to me now, but I still have this question about how the state ψ210 get unaffected by the perturbation.

    Since the perturbation is independent of z, I'm trying to find out the answer by thinking in geometry. However it is too hard for me to imagine.
  5. Aug 29, 2010 #4


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    I'm not 100% sure, but I don't think you've done the matrix elements correctly. Since both the 200 and the 210 states are rotationally symmetric to rotations in the xy plane, I think they are both unaffected by the perturbation. When I calculate the matrix elements, I find that only the W42=-W24 (in your notation) terms are non-zero.
  6. Aug 29, 2010 #5
    I've updated the given spherical harmonics for your reference.

    BTW, when calculating matrix elements, are you using polar coordinates or Cartesian coordinates?
  7. Aug 29, 2010 #6


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    I had Mathematica crank out the matrix elements. I didn't pore over it in detail, but I'm pretty sure it's correct, and it says only two matrix elements are non-zero. Attached is a pdf of the code - maybe you can see where we differ. I used spherical polar coordinates, and properly normalized wave functions.

    Attached Files:

  8. Aug 29, 2010 #7
    Yeah, you're right. I went through these elements and found that they indeed equal to zero.

    As a result, the perturbed wave function can only be a linear combination of state [tex]\psi_{21-1}[/tex] and [tex]\psi_{211}[/tex], namely the state [tex]\psi_{200}[/tex] and [tex]\psi_{210}[/tex] unaffected by the perturbation.

    Moreover, sum of first order correction of energy of state [tex]\psi_{21-1}[/tex] and [tex]\psi_{211}[/tex] is zero, so these two states will be perturbed in such a way that one of them will be lifted up and the other one lifted down with the same amount of energy.

    I'm going to work out which one up and which one down, and I guess this should be the end of the answer.

    Thanks a lot! ^_^
    Last edited: Aug 29, 2010
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